3.02f Non-uniform acceleration: using differentiation and integration

A driver is road-testing two minibuses, A and B, for a taxi company. The performance of each minibus along a straight track is compared. A flag is dropped to indicate the start of the test. Each minibus starts from rest. The acceleration in m s\(^{-2}\) of each minibus is modelled as a function of time, \(t\) seconds, after the flag is dropped: The acceleration of A = \(0.138 t^2\) The acceleration of B = \(0.024 t^3\)

1. Find the time taken for A to travel 100 metres. Give your answer to four significant figures. [4 marks]
2. The company decides to buy the minibus which travels 100 metres in the shortest time. Determine which minibus should be bought. [4 marks]
3. The models assume that both minibuses start moving immediately when \(t = 0\) In light of this, explain why the company may, in reality, make the wrong decision. [1 mark]

An elite athlete runs in a straight line to complete a 100-metre race. During the race, the athlete's velocity, \(v \text{ m s}^{-1}\), may be modelled by $$v = 11.71 - 11.68e^{-0.9t} - 0.03e^{0.3t}$$ where \(t\) is the time, in seconds, after the starting pistol is fired.

1. Find the maximum value of \(v\), giving your answer to one decimal place. Fully justify your answer. [8 marks]
2. Find an expression for the distance run in terms of \(t\). [6 marks]
3. The athlete's actual time for this race is 9.8 seconds. Comment on the accuracy of the model. [2 marks]

At time \(t = 0\), a parachutist jumps out of an airplane that is travelling horizontally. The velocity, \(\mathbf{v}\) m s\(^{-1}\), of the parachutist at time \(t\) seconds is given by: $$\mathbf{v} = (40e^{-0.2t})\mathbf{i} + 50(e^{-0.2t} - 1)\mathbf{j}$$ The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal and vertical respectively. Assume that the parachutist is at the origin when \(t = 0\) Model the parachutist as a particle.

1. Find an expression for the position vector of the parachutist at time \(t\). [4 marks]
2. The parachutist opens her parachute when she has travelled 100 metres horizontally. Find the vertical displacement of the parachutist from the origin when she opens her parachute. [4 marks]
3. Carefully, explaining the steps that you take, deduce the value of \(g\) used in the formulation of this model. [3 marks]

A particle P of mass 4 kilograms moves in such a way that its position vector at time \(t\) seconds is \(\mathbf{r}\) metres, where $$\mathbf{r} = 3t\mathbf{i} + 2e^{-3t}\mathbf{j}.$$

1. Find the initial kinetic energy of P. [4]
2. Find the time when the acceleration of P is 2 metres per second squared. [3]

At time \(t\) seconds, where \(t \geq 0\), a particle P of mass 2 kg is moving on a smooth horizontal surface. The particle moves under the action of a constant horizontal force of \((-2\mathbf{i} + 6\mathbf{j})\) N and a variable horizontal force of \((2\cos 2t \mathbf{i} + 4\sin t \mathbf{j})\) N. The acceleration of P at time \(t\) seconds is \(\mathbf{a}\) m s\(^{-2}\).

1. Find \(\mathbf{a}\) in terms of \(t\). [2]

The particle P is at rest when \(t = 0\).

1. Determine the speed of P at the instant when \(t = 2\). [5]

At time \(t\) seconds, where \(t \geq 0\), a particle P has position vector \(\mathbf{r}\) metres, where $$\mathbf{r} = (2t^2 - 12t + 6)\mathbf{i} + (t^3 + 3t^2 - 8t)\mathbf{j}.$$ The velocity of P at time \(t\) seconds is \(v \text{ m s}^{-1}\).

1. Find \(v\) in terms of \(t\). [1]
2. Determine the speed of P at the instant when it is moving parallel to the vector \(\mathbf{i} - 4\mathbf{j}\). [5]
3. Determine the value of \(t\) when the magnitude of the acceleration of P is \(20.2 \text{ m s}^{-2}\). [3]

In this question you must show detailed reasoning. In this question, positions are given relative to a fixed origin, O. The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are in the \(x\)- and \(y\)-directions respectively in a horizontal plane. Distances are measured in centimetres and the time, \(t\), is measured in seconds, where \(0 \leq t \leq 5\). A small radio-controlled toy car C moves on a horizontal surface which contains O. The acceleration of C is given by \(2\mathbf{i} + t\mathbf{j} \text{ cm s}^{-2}\). When \(t = 4\), the displacement of C from O is \(16\mathbf{i} + \frac{32}{3}\mathbf{j}\) cm, and the velocity of C is \(8\mathbf{i} \text{ cm s}^{-1}\). Determine a cartesian equation for the path of C for \(0 < t < 5\). You are not required to simplify your answer. [6]

A particle P of mass 5 kg is released from rest at a point O and falls vertically. A resistance of magnitude \(0.05v^2\) N acts vertically upwards on P, where \(v \text{ m s}^{-1}\) is the velocity of P when it has fallen a distance \(x\) m.

1. Show that \(\left(\frac{100v}{980-v^2}\right)\frac{dv}{dx} = 1\). [2]
2. Verify that \(v^2 = 980(1-e^{-0.02x})\). [4]
3. Determine the work done against the resistance while P is falling from O to the point where P's acceleration is \(8.36 \text{ m s}^{-2}\). [5]

\includegraphics{figure_7} A particle P of mass \(m\) is attached to one end of a light elastic string of natural length \(6a\) and modulus of elasticity \(3mg\). The other end of the string is fixed to a point O on a smooth plane, which is inclined at an angle of \(30°\) to the horizontal. The string lies along a line of greatest slope of the plane and P rests in equilibrium on the inclined plane at a point A, as shown in Fig. 7. P is now pulled a further distance \(2a\) down the line of greatest slope through A and released from rest. At time \(t\) later, the displacement of P from A is \(x\), where the positive direction of \(x\) is down the plane.

1. Show that, until the string slackens, \(x\) satisfies the differential equation $$\frac{d^2x}{dt^2} + \frac{gx}{2a} = 0.$$ [6]
2. Determine, in terms of \(a\) and \(g\), the time at which the string slackens. [5]
3. Find, in terms of \(a\) and \(g\), the speed of P when the string slackens. [2]

A particle moves along the horizontal \(x\)-axis so that its velocity \(v\) ms\(^{-1}\) at time \(t\) seconds is given by $$v = 6t^2 - 8t - 5.$$ At time \(t = 1\), the particle's displacement from the origin is \(-4\) m. Find an expression for the displacement of the particle at time \(t\) seconds. [3]

A particle \(P\), of mass 3 kg, moves along the horizontal \(x\)-axis under the action of a resultant force \(F\) N. Its velocity \(v\) ms\(^{-1}\) at time \(t\) seconds is given by $$v = 12t - 3t^2.$$

1. Given that the particle is at the origin \(O\) when \(t = 1\), find an expression for the displacement of the particle from \(O\) at time \(t\) s. [3]
2. Find an expression for the acceleration of the particle at time \(t\) s. [2]

The position vector \(\mathbf{x}\) metres at time \(t\) seconds of an object of mass 3 kg may be modelled by $$\mathbf{x} = 3\sin t \mathbf{i} - 4\cos 2t \mathbf{j} + 5\sin t \mathbf{k}.$$

1. Find an expression for the velocity vector \(\mathbf{v}\text{ ms}^{-1}\) at time \(t\) seconds and determine the least value of \(t\) when the object is instantaneously at rest. [7]
2. Write down the momentum vector at time \(t\) seconds. [1]
3. Find, in vector form, an expression for the force acting on the object at time \(t\) seconds. [3]

A particle \(P\) moves in a straight line. At time \(t\) s the displacement \(s\) cm from a fixed point \(O\) is given by: $$s = \frac{1}{6}\left(8t^3 - 105t^2 + 144t + 540\right).$$ Find the distance between the points at which the particle is instantaneously at rest. [7]

A car starts from the point \(A\). At time \(t\) s after leaving \(A\), the distance of the car from \(A\) is \(s\) m, where \(s = 30t - 0.4t^2\), \(0 \leq t \leq 25\). The car reaches the point \(B\) when \(t = 25\).

1. Find the distance \(AB\). [2]
2. Show that the car travels with a constant acceleration and state the value of this acceleration. [3]

A runner passes through \(B\) when \(t = 0\) with an initial velocity of \(2 \text{ m s}^{-1}\) running directly towards \(A\). The runner has a constant acceleration of \(0.1 \text{ m s}^{-2}\).

1. Find the distance from \(A\) at which the runner and the car pass one another. [8]

Two particles \(A\) and \(B\) have position vectors \(\mathbf{r}_A\) metres and \(\mathbf{r}_B\) metres at time \(t\) seconds, where $$\mathbf{r}_A = t^2\mathbf{i} + (3t - 1)\mathbf{j} \quad \text{and} \quad \mathbf{r}_B = (1 - 2t^2)\mathbf{i} + (3t - 2t^2)\mathbf{j}, \quad \text{for } t \geqslant 0.$$

1. Find the values of \(t\) when \(A\) and \(B\) are moving with the same speed. [5]
2. Show that the distance, \(d\) metres, between \(A\) and \(B\) at time \(t\) satisfies $$d^2 = 13t^4 - 10t^2 + 2.$$ [3]
3. Hence find the shortest distance between \(A\) and \(B\) in the subsequent motion. [6]

A particle \(P\) moves along the \(x\)-axis. At time \(t\) seconds the velocity of \(P\) is \(v\text{m s}^{-1}\), where \(v = 2t^4 + kt^2 - 4\). The acceleration of \(P\) when \(t = 2\) is \(28\text{m s}^{-2}\).

1. Show that \(k = -9\). [3]
2. Show that the velocity of \(P\) has its minimum value when \(t = 1.5\). [3]

When \(t = 1\), \(P\) is at the point \((-6.4125, 0)\).

1. Find the distance of \(P\) from the origin \(O\) when \(P\) is moving with minimum velocity. [4]

A student is attempting to model the flight of a boomerang. She throws the boomerang from a fixed point \(O\) and catches it when it returns to \(O\). She suggests the model for the displacement, \(s\) metres, after \(t\) seconds is given by \(s = 9t^2 - \frac{3}{2}t^3\), \(0 \leq t \leq 6\). For this model,

1. determine what happens at \(t = 6\), [2]
2. find the greatest displacement of the boomerang from \(O\), [4]
3. find the velocity of the boomerang 1 second before the student catches it, [2]
4. find the acceleration of the boomerang 1 second before the student catches it. [2]

A particle moves along a straight line under the action of a variable force. The acceleration is given by $$a = \begin{cases} 30 - 6t, & \text{for } 0 \leqslant t \leqslant 10 \\ 6t - 90, & \text{for } 10 \leqslant t \leqslant 20 \end{cases}$$ where time \(t\) is measured in seconds and \(a\) in m s\(^{-2}\). The particle is at rest at the origin at \(t = 0\).

1. 1. Find the velocity \(v\) of the particle in terms of \(t\). Verify that \(v = 0\) when \(t = 10\) and \(t = 20\). [7]
   2. Sketch the velocity-time graph for the motion. [2]
2. Calculate the total distance travelled by the particle. [4]

A particle \(P\) is free to move along a straight line \(Ox\). It starts from rest at \(O\) and after \(t\) seconds its acceleration \(a\,\mathrm{m}\,\mathrm{s}^{-2}\) is given by \(a = 12 - 6t\).

1. Find an expression in terms of \(t\) for its velocity \(v\,\mathrm{m}\,\mathrm{s}^{-1}\). Hence find the velocity of \(P\) when \(t = 4\). [4]
2. Find the displacement of \(P\) from \(O\) when \(t = 4\). [3]
3. Find the velocity of \(P\) when it returns to \(O\). [3]

An engine is travelling along a straight horizontal track against negligible resistances. In travelling a distance of 750 m its speed increases from 5 m s\(^{-1}\) to 15 m s\(^{-1}\). Find the time taken if the engine was

1. exerting a constant tractive force, [2]
2. working at constant power. [9]

A particle \(P\) is free to move along a straight line \(Ox\). It starts from rest at \(O\) and after \(t\) seconds its acceleration \(a \mathrm{~m} \mathrm{~s}^{-2}\) is given by \(a = 12 - 6t\).

1. Find an expression in terms of \(t\) for its velocity \(v \mathrm{~m} \mathrm{~s}^{-1}\). Hence find the velocity of \(P\) when \(t = 4\). [4]
2. Find the displacement of \(P\) from \(O\) when \(t = 4\). [3]
3. Find the velocity of \(P\) when it returns to \(O\). [3]

A particle \(P\) is free to move along a straight line \(Ox\). It starts from rest at \(O\) and after \(t\) seconds its acceleration \(a \mathrm{m} \mathrm{s}^{-2}\) is given by \(a = 12 - 6t\).

1. Find an expression in terms of \(t\) for its velocity \(v \mathrm{m} \mathrm{s}^{-1}\). Hence find the velocity of \(P\) when \(t = 4\). [4]
2. Find the displacement of \(P\) from \(O\) when \(t = 4\). [3]
3. Find the velocity of \(P\) when it returns to \(O\). [3]

A particle of mass \(0.01\) kg is projected vertically upwards from a point \(G\) at ground level with speed \(165 \text{ m s}^{-1}\) and reaches a maximum height of \(1237.5\) m. Throughout its motion it experiences a constant resistance.

1. Find the acceleration of the particle as it ascends and hence the magnitude of the resistance. [4]
2. During its descent back to \(G\) the particle experiences the same constant resistance. Find the time taken for the descent. [4]

7 A small ball \(B\) of mass 0.2 kg moves in a narrow fixed smooth cylindrical tube \(O A\) of length 1 m , closed at the end \(A\). When the ball has displacement \(x \mathrm {~m}\) from \(O\), it has velocity \(v \mathrm {~ms} ^ { - 1 }\) in the direction \(O A\) and experiences a resisting force of magnitude \(\frac { k } { 1 - x } \mathrm {~N}\).

1. \includegraphics[max width=\textwidth, alt={}, center]{10abedc3-c814-47c0-8ed4-849ef325feca-4_186_805_488_715} The tube is fixed in a horizontal position and \(B\) is projected from \(O\) towards \(A\) with velocity \(1.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) (see diagram). Given that \(B\) comes to instantaneous rest after travelling 0.55 m , show that \(k = 0.1803\), correct to 4 significant figures.
2. The tube is now fixed in a vertical position with \(O\) above \(A\). The ball \(B\) is released from rest at \(O\). Calculate the speed of \(B\) after it has descended 0.1 m . \end{document}

4 A particle \(P\) moves on a straight line, starting from rest at a point \(O\) of the line. The time after \(P\) starts to move is \(t \mathrm {~s}\), and the particle moves along the line with constant acceleration \(\frac { 1 } { 4 } \mathrm {~m} \mathrm {~s} ^ { - 2 }\) until it passes through a point \(A\) at time \(t = 8\). After passing through \(A\) the velocity of \(P\) is \(\frac { 1 } { 2 } t ^ { \frac { 2 } { 3 } } \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Find the acceleration of \(P\) immediately after it passes through \(A\). Hence show that the acceleration of \(P\) decreases by \(\frac { 1 } { 12 } \mathrm {~m} \mathrm {~s} ^ { - 2 }\) as it passes through \(A\).
2. Find the distance moved by \(P\) from \(t = 0\) to \(t = 27\).